Intro Predictive Maintenance 2 Part 5 pot

broadband, narrowband, or component. The unit of measurement is useful when the analyst needs to know the total displacement or maximum energy produced by the machine’s vibration profile. Technically, peak-to-peak values should be used in conjunction with actual shaft- displacement data, which are measured with a proximity or displacement transducer. Peak-to-peak terms should not be used for vibration data acquired using either relative vibration data from bearing caps or when using a velocity or acceleration transducer. The only exception is when vibration levels must be compared to vibra- tion-severity charts based on peak-to-peak values. Zero-to-peak. Zero-to-peak (A), or simply peak, values are equal to one half of the peak-to-peak value. In general, relative vibration data acquired using a velocity trans- ducer are expressed in terms of peak. Root-mean-square. Root-mean-square (RMS) is the statistical average value of the amplitude generated by a machine, one of its components, or a group of components. Referring to Figure 7–11, RMS is equal to 0.707 of the zero-to-peak value, A. Nor- mally, RMS data are used in conjunction with relative vibration data acquired using an accelerometer or expressed in terms of acceleration. 7.5 M ACHINE DYNAMICS The primary reasons for vibration-profile variations are the dynamics of the machine, which are affected by mass, stiffness, damping, and degrees of freedom; however, care 132 An Introduction to Predictive Maintenance Figure 7–11 Relationship of vibration amplitude. must be taken because the vibration profile and energy levels generated by a machine may vary depending on the location and orientation of the measurement. 7.5.1 Mass, Stiffness, and Damping The three primary factors that determine the normal vibration energy levels and the resulting vibration profiles are mass, stiffness, and damping. Every machine-train is designed with a dynamic support system that is based on the following: the mass of the dynamic component(s), specific support system stiffness, and a specific amount of damping. Mass Mass is the property that describes how much material is present. Dynamically, the property describes how an unrestricted body resists the application of an external force. Simply stated, the greater the mass, the greater the force required to accelerate it. Mass is obtained by dividing the weight of a body (e.g., rotor assembly) by the local acceleration of gravity, g. The English system of units is complicated compared to the metric system. In the English system, the units of mass are pounds-mass (lbm) and the units of weight are pounds-force (lbf). By definition, a weight (i.e., force) of one lbf equals the force pro- duced by one lbm under the acceleration of gravity. Therefore, the constant, g c , which has the same numerical value as g (32.17) and units of lbm-ft/lbf-sec 2 , is used in the definition of weight: Therefore, Therefore, Stiffness Stiffness is a spring-like property that describes the level of resisting force that results when a body changes in length. Units of stiffness are often given as pounds per inch Mass Weight lbf ft lbm ft lbf lbm==¥= * sec * *sec g g c 2 2 Mass Weight = * g g c Weight Mass = * g g c Vibration Monitoring and Analysis 133 (lbf/in). Machine-trains have three stiffness properties that must be considered in vibration analysis: shaft stiffness, vertical stiffness, and horizontal stiffness. Shaft Stiffness. Most machine-trains used in industry have flexible shafts and rela- tively long spans between bearing-support points. As a result, these shafts tend to flex in normal operation. Three factors determine the amount of flex and mode shape that these shafts have in normal operation: shaft diameter, shaft material properties, and span length. A small-diameter shaft with a long span will obviously flex more than one with a larger diameter or shorter span. Vertical Stiffness. The rotor-bearing support structure of a machine typically has more stiffness in the vertical plane than in the horizontal plane. Generally, the structural rigidity of a bearing-support structure is much greater in the vertical plane. The full weight of and the dynamic forces generated by the rotating element are fully sup- ported by a pedestal cross-section that provides maximum stiffness. In typical rotating machinery, the vibration profile generated by a normal machine contains lower amplitudes in the vertical plane. In most cases, this lower profile can be directly attributed to the difference in stiffness of the vertical plane when compared to the horizontal plane. Horizontal Stiffness. Most bearing pedestals have more freedom in the horizontal direction than in the vertical. In most applications, the vertical height of the pedestal is much greater than the horizontal cross-section. As a result, the entire pedestal can flex in the horizontal plane as the machine rotates. This lower stiffness generally results in higher vibration levels in the horizontal plane. This is especially true when the machine is subjected to abnormal modes of operation or when the machine is unbalanced or misaligned. Damping Damping is a means of reducing velocity through resistance to motion, in particular by forcing an object through a liquid or gas, or along another body. Units of damping are often given as pounds per inch per second (lbf/in/sec, which is also expressed as lbf-sec/in). The boundary conditions established by the machine design determine the freedom of movement permitted within the machine-train. A basic understanding of this concept is essential for vibration analysis. Free vibration refers to the vibration of a damped (as well as undamped) system of masses with motion entirely influenced by their potential energy. Forced vibration occurs when motion is sustained or driven by an applied periodic force in either damped or undamped systems. The following sections discuss free and forced vibration for both damped and undamped systems. Free Vibration—Undamped. To understand the interactions of mass and stiffness, consider the case of undamped free vibration of a single mass that only moves 134 An Introduction to Predictive Maintenance vertically, which is illustrated in Figure 7–12. In this figure, the mass “M” is sup- ported by a spring that has a stiffness “K” (also referred to as the spring constant), which is defined as the number of pounds tension necessary to extend the spring one inch. The force created by the static deflection, X i , of the spring supports the weight, W, of the mass. Also included in Figure 7–12 is the free-body diagram that illustrates the two forces acting on the mass. These forces are the weight (also referred to as the inertia force) and an equal, yet opposite force that results from the spring (referred to as the spring force, F s ). The relationship between the weight of mass, M, and the static deflection of the spring can be calculated using the following equation: W = KX i If the spring is displaced downward some distance, X 0 , from X i and released, it will oscillate up and down. The force from the spring, F s , can be written as follows, where “a” is the acceleration of the mass: It is common practice to replace acceleration, a, with the second derivative of the displacement, X, of the mass with respect to time, t. Making this substitution, the equation that defines the motion of the mass can be expressed as: Motion of the mass is known to be periodic. Therefore, the displacement can be described by the expression: M g dX dt KX or M g dX dt KX cc 2 2 2 2 0=- + = dX dt 2 2 , FKX Ma g s c =- = Vibration Monitoring and Analysis 135 Mass Spring Mass Weight (W ) F s Static Deflection (X ) Figure 7–12 Undamped spring-mass system. Where: X = Displacement at time t X 0 = Initial displacement of the mass w = Frequency of the oscillation (natural or resonant frequency) t = Time If this equation is differentiated and the result inserted into the equation that defines motion, the natural frequency of the mass can be calculated. The first derivative of the equation for motion yields the equation for velocity. The second derivative of the equation yields acceleration. Inserting the expression for acceleration, or into the equation for F s yields the following: Solving this expression for w yields the equation: Where: w = Natural frequency of mass K = Spring constant M = Mass Note that, theoretically, undamped free vibration persists forever; however, this never occurs in nature, and all free vibrations die down after time because of damping, which is discussed in the next section. w Kg M c = M g dX dt KX M g XtKX M g XKX M g K c c cc 2 2 2 0 22 0 0 0 += - () += -+=-+= ww ww cos dX dt 2 2 , Velocity dX dt XX t Acceleration dX dt XXt ===- () ===- () ˙ sin ˙˙ cos ww ww 0 2 2 2 0 XX t= () 0 cos w 136 An Introduction to Predictive Maintenance Free Vibration—Damped. A slight increase in system complexity results when a damping element is added to the spring-mass system shown in Figure 7–13. This type of damping is referred to as viscous damping. Dynamically, this system is the same as the undamped system illustrated in Figure 7–12, except for the damper, which usually is an oil or air dashpot mechanism. A damper is used to continuously decrease the velocity and the resulting energy of a mass undergoing oscillatory motion. The system consists of the inertia force caused by the mass and the spring force, but a new force is introduced. This force is referred to as the damping force and is pro- portional to the damping constant, or the coefficient of viscous damping, c. The damping force is also proportional to the velocity of the body and, as it is applied, it opposes the motion at each instant. In Figure 7–13, the nonelongated length of the spring is “L o ” and the elongation caused by the weight of the mass is expressed by “h.” Therefore, the weight of the mass is Kh. Part (a) of Figure 7–13 shows the mass in its position of stable equilibrium. Part (b) shows the mass displaced downward a distance X from the equilibrium position. Note that X is considered positive in the downward direction. Part (c) of Figure 7–13 is a free-body diagram of the mass, which has three forces acting on it. The weight (Mg/g c ), which is directed downward, is always positive. The damping force which is the damping constant times velocity, acts opposite to the direction of the velocity. The spring force, K(X + h), acts in the direction opposite c dX dt Ê Ë ˆ ¯ , Vibration Monitoring and Analysis 137 Figure 7–13 Damped spring-mass system. to the displacement. Using Newton’s equation of motion, where SF = Ma, the sum of the forces acting on the mass can be represented by the following equation, remem- bering that X is positive in the downward direction: Dividing by In order to look up the solution to the above equation in a differential equations table (such as in CRC Handbook of Chemistry and Physics), it is necessary to change the form of this equation. This can be accomplished by defining the relationships, cg c /M = 2m and Kg c /M = w 2 , which converts the equation to the following form: Note that for undamped free vibration, the damping constant, c, is zero and, therefore, m is zero. The solution of this equation describes simple harmonic motion, which is given as follows: Substituting at t = 0, then X = X 0 and then X = X 0 cos(wt) This shows that free vibration is periodic and is the solution for X. For damped free vibration, however, the damping constant, c, is not zero. dX dt = 0, XA t B t= () + () cos sinww dX dt X dX dt X 2 2 2 2 2 2 0 =- =+ = w w dX dt dX dt X 2 2 2 2=- -mw dX dt cg M dX dt Kg X M cc 2 2 =- - M g c : M g dX dt Mg g c dX dt KX h M g dX dt Kh c dX dt KX Kh M g dX dt c dX dt KX cc c c 2 2 2 2 2 2 =- -+ () =- =- - 138 An Introduction to Predictive Maintenance or or D 2 + 2mD + w 2 = 0 which has a solution of: X = Ae d 1 t + B e d 2 t where: There are different conditions of damping: critical, overdamping, and underdamping. Critical damping occurs when m equals w. Overdamping occurs when m is greater than w. Underdamping occurs when m is less than w. The only condition that results in oscillatory motion and, therefore, represents a mechanical vibration is underdamping. The other two conditions result in periodic motions. When damping is less than critical (m < w), then the following equation applies: where: Forced Vibration—Undamped. The simple systems described in the preceding two sections on free vibration are alike in that they are not forced to vibrate by any excit- ing force or motion. Their major contribution to the discussion of vibration funda- mentals is that they illustrate how a system’s natural or resonant frequency depends on the mass, stiffness, and damping characteristics. The mass-stiffness-damping system also can be disturbed by a periodic variation of external forces applied to the mass at any frequency. The system shown in Figure 7–12 is increased in complexity by adding an external force, F 0 , acting downward on the mass. a w m 1 22 = - X X ett t =+ ( ) - 0 1 11 1 a aama m cos sin d d 1 22 2 22 =- + - =- - - m mw m m w dX dt dX dt X 2 2 2 20++=mw dX dt dX dt X 2 2 2 2=- -mw Vibration Monitoring and Analysis 139 In undamped forced vibration, the only difference in the equation for undamped free vibration is that instead of the equation being equal to zero, it is equal to F 0 sin(wt): Because the spring is not initially displaced and is “driven” by the function F 0 sin(wt), a particular solution, X = X 0 sin(wt), is logical. Substituting this solution into the above equation and performing mathematical manipulations yields the following equation for X: where: X = Spring displacement at time, t X st = Static spring deflection under constant load, F 0 w = Forced frequency w n = Natural frequency of the oscillation t = Time C 1 and C 2 = Integration constants determined from specific boundary conditions In the above equation, the first two terms are the undamped free vibration, whereas the third term is the undamped forced vibration. The solution, containing the sum of two sine waves of different frequencies, is not a harmonic motion. Forced Vibration—Damped. In a damped forced vibration system such as the one shown in Figure 7–14, the motion of the mass “M” has two parts: (1) the damped free vibration at the damped natural frequency and (2) the steady-state harmonic motions at the forcing frequency. The damped natural frequency component decays quickly, but the steady-state harmonic associated with the external force remains as long as the energy force is present. With damped forced vibration, the only difference in its equation and the equation for damped free vibration is that it is equal to F 0 sin(wt) as shown below instead of being equal to zero. With damped vibration, the damping constant, “c,” is not equal to zero and the solu- tion of the equation becomes complex assuming the function, X = X 0 sin(wt - f). In M g dX dt c dX dt KX F t c 2 2 0 ++= () sin w XC t C t X t nn st n = () + () + - () () 12 2 1 sin cos sinww ww w M g dX dt KX F t c 2 2 0 += () sin w 140 An Introduction to Predictive Maintenance this equation, f is the phase angle, or the number of degrees that the external force, F 0 sin(wt), is ahead of the displacement, X 0 sin(wt - f). Using vector concepts, the fol- lowing equations apply, which can be solved because there are two equations and two unknowns: Solving these two equations for the unknowns X 0 and f: Where: c = Damping constant c/c c = Damping ratio F 0 = External force F 0 /K = Deflection of the spring under load, F 0 (also called static deflection, X st ) w = Forced frequency w n = Natural frequency of the oscillation w/w n = Frequency ratio c M g c c n ==Critical damping 2 w X F cK M g F K c c c K M g c c o c ncn c cn n = () +- Ê Ë Á ˆ ¯ ˜ = - Ê Ë ˆ ¯ +¥ Ê Ë ˆ ¯ = - = ¥ - () 0 2 2 2 0 2 2 22 2 22 12 2 1 ww w w w w f w w w w ww tan Vertical vector component: Horizontal vector com p onent: KX M g XF cX F c 0 2 00 00 0 0 = -= wf w f cos sin Vibration Monitoring and Analysis 141 Mass Mass Spring –KX –C dX dt F 0 Sin (wt ) Figure 7–14 Damped forced vibration system. [...]... 7 2 These data are applicable for comparison with filtered narrowband data taken from machine-trains 164 An Introduction to Predictive Maintenance Table 7 2 Vibration-Severity Standards (Inches/Second-Peak) Machine Classes Condition Good Operating Condition Alert Limit Alarm Limit Absolute Fault Limit I II III IV 0. 028 0.010 0. 156 0 .26 0 0.0 42 0. 156 0.396 0.400 0.100 0 . 25 5 0.396 0. 620 0. 156 0.396 0. 622 ... amplitudes, A1 and A2, their behavior can be represented as: X1 = A1 sin(wt ) X2 = A2 sin(wt ) By substituting these into the differential equations, two equations for the amplitude A1 ratio, , can be found: A2 A1 - K3 = M1 2 A2 w - K1 - K3 gc and M2 2 w - K 2 - K3 A1 g = c A2 - K3 For a solution of the form we assumed to exist, these two equations must be equal: M2 2 w - K 2 - K3 gc = M1 2 - K3 w - K1 -... K3 K 2 + K3 ¸ K1 K 2 + K 2 K3 + K1 K3 w4 -w2Ì + =0 ˝+ M1 M2 M2 gc ˛ Ó M1 gc 2 gc This equation, known as the frequency equation, has two solutions for w2 When substituted in either of the preceding equations, each one of these gives a definite value 146 An Introduction to Predictive Maintenance A1 This means that there are two solutions for this example, which are of the form A2 A1 sin(wt) and A2 sin(wt)... equation of motion for the second mass, M2, is derived in the same manner To make it easier to understand, turn the figure upside down and reverse the direction of X1 and X2 The equation then becomes: Vibration Monitoring and Analysis 1 45 M2 ˙˙ X2 = - K2 X2 + K3 ( X1 - X2 ) gc or M2 ˙˙ X2 + ( K2 + K3 ) X2 - K3 X1 = 0 gc If we assume that the masses, M1 and M2, undergo harmonic motions with the same... proximity probes is cost The typical cost for installing a single probe, including a power supply, signal conditioning, and so on, 154 An Introduction to Predictive Maintenance 1 6 2 3 5 4 SENSITIVE AXIS (1) Pickup case (2) Wire out (3) Damper (4) Mass (5) Spring (6) Magnet Figure 7 22 Schematic diagram of velocity pickup averages $1,000 If each machine to be evaluated requires 10 measurements, the cost per... f(t), can be represented as a series of sine functions having frequencies w, 2w, 3w, 4w, and so on Function f(t) is represented by the following equation, which is referred to as a Fourier Series: f (t ) = A0 + A1 sin(wt + f1 ) + A2 sin(2wt + f 2 ) + A3 sin(3wt + f 3 ) + 150 An Introduction to Predictive Maintenance Figure 7 20 Relationship between time-domain and frequency-domain where: Ax = Amplitude... undamped two-degree-of-freedom system is illustrated in Figure 7–16 This diagram consists of two masses, M1 and M2, that are suspended from springs, K1 and K2 The two masses are tied together, or coupled, by spring, K3, so that they are 144 An Introduction to Predictive Maintenance k1 M1 X1 k3 M2 X2 k2 Figure 7–16 Undamped two-degreesof-freedom system with a spring couple forced to act together In this example,...1 42 An Introduction to Predictive Maintenance For damped forced vibrations, three different frequencies have to be distinguished: the undamped natural frequency, Kgc M ; the damped natural frequency, 2 Kgc Ê cgc ˆ q= ; and the frequency of maximum forced amplitude, sometimes M Ë 2M ¯ referred to as the resonant frequency 7 .5 .2 Degrees of Freedom In a mechanical system,... 0.0 05 seconds The impact is clearly visible in both the vertical (top) and horizontal (bottom) data set 148 An Introduction to Predictive Maintenance From these time traces, the vertical impact appears to be stronger than the horizontal In addition, the impact repeated at 0.0 15 and 0. 0 25 seconds Two conclusions can be derived from this example: (1) the impact source is a vertical force, and (2) it... faulty readings Figure 7 28 Handheld transducers should be avoided when possible maximum benefit from a predictive maintenance program If this technique must be used, extreme care should be exercised to ensure that the same location, orientation, and compressive load are used for every measurement Figure 7 28 illustrates a handheld device 160 An Introduction to Predictive Maintenance 7.7.4 Acquiring . - K M g KK M g KK K c c 3 1 2 13 2 2 23 3 w w A A M g KK K c 1 2 2 2 23 3 = - w A A K M g KK c 1 2 3 1 2 13 = - w A A 1 2 , XA t XA t 11 22 = () = () sin sin w w M g XKKXKX c 2 223 231 0 ˙˙ ++ () -= M g XKXKXX c 2 222 3 12 ˙˙ =-. sinww dX dt X dX dt X 2 2 2 2 2 2 0 =- =+ = w w dX dt dX dt X 2 2 2 2=- -mw dX dt cg M dX dt Kg X M cc 2 2 =- - M g c : M g dX dt Mg g c dX dt KX h M g dX dt Kh c dX dt KX Kh M g dX dt c dX dt KX cc c c 2 2 2 2 2 2 =-. ratio c M g c c n ==Critical damping 2 w X F cK M g F K c c c K M g c c o c ncn c cn n = () +- Ê Ë Á ˆ ¯ ˜ = - Ê Ë ˆ ¯ +¥ Ê Ë ˆ ¯ = - = ¥ - () 0 2 2 2 0 2 2 22 2 22 12 2 1 ww w w w w f w w w w ww tan Vertical